Question

我想从函数中返回一个有区别的联合类型 - 它与类型推断冲突 - 我应该如何更改我的代码以便getKeyA返回KeyA变为Key？

      type KeyA = {keyString:string}
type KeyB = {keyInt:int}
type Key = KeyA | KeyB

let getKeyA id =
         {keyString="123"}
let getKeyB id = 
         {keyInt=2}

let getKey (id) :Key = 
     match id with 
      | 1 -> getKeyA id 
      | _ -> getKeyB id

Answer 1

简而言之：我不知道您提供的大多数部分的定义，但要返回Key类型的内容，您必须在此示例中使用KeyA或KeyB

试试这个：

type Key = KeyA of KeyA | KeyB of KeyB

let getKey id : Key = 
    match id with 
    | 1 -> KeyA (getKeyA id)
    | _ -> KeyB (getKeyB id)

或者这看起来更适合你

type KeyA = {keyString:string}
type KeyB = {keyInt:int}
type Key = KeyA of KeyA | KeyB of KeyB

let getKeyA id =
    { keyString="123" }
    |> KeyA

let getKeyB id = 
    { keyInt=2 }
    |> KeyB

let getKey id : Key = 
     match id with 
      | 1 -> getKeyA id 
      | _ -> getKeyB id

您可能会注意到，您只将string或int包装到记录中 - 如果它只是一种类型/值，或者如果您可以使用元组，我会删除其他记录：< / p>

type Key = KeyA of string | KeyB of int

let getKeyA id = KeyA "123"

let getKeyB id = KeyB 2

let getKey id : Key = 
     match id with 
      | 1 -> getKeyA id 
      | _ -> getKeyB id

F＃使用受歧视的联盟类型

1 个答案: